In the case of a real valued matrix the existence of the Jordan form is not guaranteed, because the real numbers are not algebraically closed (i.e. the characteristic polynomial might not split), so I do not think you have provided enough reason why in this particular case the Jordan form exists.
I would suggest that you first show this is so by proving that all the characteristic values must be real, as proved in another answer here...then you can assert that $A$ does indeed have a Jordan form.
Now I can see what you are trying to say with the Jordan form: $J^T$ is also a Jordan canonical form matrix - some textbooks use the super diagonal and some the sub diagonal - and they are similar to one another. Let's say a matrix $A$ is similar to $J$ with 1's in the super diagonal, and then it would also be similar to $J^T$ - now the ones are in the sub diagonal. The "uniqueness" of the Jordan form is a tricky argument here though - $J$ is unique and $J^T$ is also unique - it does not mean they are equal though. I think if you want to go this root you should try to prove with an algebraic argument that $J=J^T$ explicitly. Then $J$ is a diagonal matrix, which is what we want, however you don't prove this ...
Lastly, the standard way to show that a symmetric matrix is similar to a diagonal matrix is to make use of Schur's theorem - it is an easier route in IMHO: if all the characteristic roots of a matrix $A$ is real (second paragraph!) then there is an orthogonal matrix $P$ so that $P^TAP$ is upper triangular...now it is easy to verify that $P^TAP$ is symmetric and therefore diagonal, and so even more can be said: $A$ is orthogonally diagonalizable.